Question

A coil of 30 turns is wrapped around a long solenoid of cross-sectional area 7.0 ✕ 10−3 m2. The solenoid is 0.40 m long and has 600 turns. (a) What is the mutual inductance of this system (in H)? H (b) The outer coil is replaced by a coil of 30 turns whose radius is three times that of the solenoid. What is the mutual inductance of this configuration (in H)?

Answer 1

a) 1.26×10⁻⁴ H

b) 1.26×10⁻⁴ H

Explanation:

N₁ = Number of turns in the solenoid = 600

N₂ = Number of turns in the coil = 30

A = Cross sectional area = 7×10⁻³ m²

l = Length of the solenoid = 0.4 m

μ₀ = Permeability of free space = 4π10⁻⁷ T m/A

a) Mutual inductance

`M=(\mu_(0) N_1N_2A)/(l)\\\Rightarrow M=(4* 10^(-7)* 600* 30* 7* 10^(-3))/(0.4)\\\Rightarrow M=1.26* 10^(-4)\ H`

∴ Mutual inductance of this system is 1.26×10⁻⁴ H

b) Mutual inductance does not depend on the cross sectional of the outer coil but depends on the area of the solenoid.

Hence, the mutual inductance remains the same.